We know that when the polarizers are rotated for angle θ, the transmitted intensity will be
It=I0 cos2θ.This is all we need to solve the problem.
a. At 90° angle the percentage of the initial intensity transmitted is:
I0It= cos290∘⋅100%=0.b. After the first polarizer we only have a half of the initial intensity:
I1=0.5I0.
After the first shift (after two sheets 45° to each other, before the third polarizer) we have
I2=I1 cos245∘=0.5I1=0.25I0.After the second shift (after the third polarizer):
I3=I2 cos245∘=0.5I2=0.125I0,
or 12.5%.
c. According to the previous case, we have half initial intensity after the first sheet,
I2=I1 cos230∘=0.75I1=0.375I0 after the second sheet,
I3=I2 cos230∘=0.75I2=0.28I0 after the third one,
I4=I3 cos230∘=0.75I3=0.21I0after the four sheets shifted 30° to each other, we have 21% of the initial intensity.
d. From problems (b) and (c) we could see the pattern. After 3 and 4 sheets the final intensity respectively was
I3=0.5I0[cos2θ]3−1,I4=0.5I0[cos2θ]4−1,therefore, the rule is
In=2I0[cos2θ]n−1. Since we need 90° shift, angle θ must comply with the rule
90∘=(n−1)θ, therefore, the transmitted intensity is
In=I02[cos2(n−190∘)]n−1.
To keep 90% intensity, we must find n from the following equation:
I0In=0.9=2[cos2(n−190∘)]n−1. This equation does not have solution.
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